Q133 / 180 · KCET 2015 · Physics
The ratio of kinetic energy to the potential energy of a particle executing SHM at a distance
equal to half its amplitude, the distance being measured from its equilibrium position is
equal to half its amplitude, the distance being measured from its equilibrium position is
- Correct answerA\( 3: 1 \)
- B\( 04: 1 \)
- C(1) \( 1 \)
- D\( 8: 1 \)
Solution
Kinetic energy of particle executing SHM is
\( K E=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right) \)
Hey \( y=\frac{A}{2} \)
\( \therefore K E=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)=\frac{1}{2} m \omega^{2} \frac{3 A^{2}}{4} \)
Potential energy of particle executing SHM is
\( P E=\frac{1}{2} m \omega^{2} y^{2}=\frac{1}{2} m \omega^{2} \frac{A^{2}}{4} \)
Then ratio of kinetic energy to potential energy is
\( \frac{K E}{P E}=\frac{\left(\frac{1}{2}\right) m \omega^{2}\left(\frac{3}{4}\right) A^{2}}{\left(\frac{1}{2}\right) m \omega^{2} \frac{A^{2}}{4}}=\frac{3}{1} \)
Therefore, the ratio is \( 3: 1 \).
\( K E=\frac{1}{2} m \omega^{2}\left(A^{2}-y^{2}\right) \)
Hey \( y=\frac{A}{2} \)
\( \therefore K E=\frac{1}{2} m \omega^{2}\left(A^{2}-\frac{A^{2}}{4}\right)=\frac{1}{2} m \omega^{2} \frac{3 A^{2}}{4} \)
Potential energy of particle executing SHM is
\( P E=\frac{1}{2} m \omega^{2} y^{2}=\frac{1}{2} m \omega^{2} \frac{A^{2}}{4} \)
Then ratio of kinetic energy to potential energy is
\( \frac{K E}{P E}=\frac{\left(\frac{1}{2}\right) m \omega^{2}\left(\frac{3}{4}\right) A^{2}}{\left(\frac{1}{2}\right) m \omega^{2} \frac{A^{2}}{4}}=\frac{3}{1} \)
Therefore, the ratio is \( 3: 1 \).
Hence, the correct answer is option: 'A'.